# General topology / real analysis - Suppose $S$ is a bounded and closed nonempty subset of real numbers. Prove sup S is in S

Q. Suppose S is a bounded and closed nonempty subset of real numbers. Prove $$\sup S$$ is in $$S$$.

Since $$S$$ is bounded and by the least upper bound property of $$\mathbb R$$, there exists $$\sup S \in \mathbb R$$ which we do not know if it is in $$S$$ yet.

By the definition of supremum, for $$\epsilon \gt 0$$,

$$\exists s \in S$$ such that, $$\sup S - \epsilon \lt s \lt \sup S + \epsilon$$

So $$s$$ is in the interval, $$(\sup S - \epsilon, \sup S + \epsilon)$$

Then $$S \cap (\sup S - \epsilon, \sup S + \epsilon) \neq \emptyset$$

Then $$\Bigl( (\sup S - \epsilon, \sup S + \epsilon) \setminus \{\sup S\} \Bigr) \cap S \neq \emptyset$$

We also know that a closed set contains all of its accumulation points and since $$\sup S$$ is an accumulation point of $$S$$, we proved $$\sup S \in S$$

Does this proof look okay? If not, can you give me hints on where I went wrong?

Thank you.

It is almost fine, but not entirely. You jumped from$$S\cap(\sup S-\varepsilon,\sup S+\varepsilon)\neq\emptyset$$to$$S\cap\bigl((\sup S-\varepsilon,\sup S+\varepsilon)\setminus\{\sup S\}\bigr)\neq\emptyset\tag1$$without any justification. If it turns out that $$s=\sup S$$, $$(1)$$ may well be false.

This is easy to solve, though. Just begin by saying that this is a proof by contradiction. That is, you assume that $$\sup S\notin S$$. Then $$(1)$$ would be true and you reach a contradiction when you arive that $$\sup S\in S$$ after all.

• I'm quite confused. Can you elaborate more on if I assume that sup $S \notin S$, then (1) would be true? thank you – TUC Oct 28 '18 at 9:52
• You know that $S\cap(\sup S-\varepsilon,\sup S+\varepsilon)\neq\emptyset$ because you proved that there is some $s\in S\cap(\sup S-\varepsilon,\sup S+\varepsilon)$. But if that $s$ is precisely $\sup S$, then how do you know that $(1)$ holds? The only difference between $S\cap(\sup S-\varepsilon,\sup S+\varepsilon)$ and $S\cap\bigl((\sup S-\varepsilon,\sup S+\varepsilon)\setminus\{\sup S\}\bigr)$ is that $\sup S$ belongs to the first set, but not to the second one. – José Carlos Santos Oct 28 '18 at 10:00

At the beginning, you better mention that your S is (non-empty) and bounded so there is a real supremum.

• Every closed and bounded set $S$ in $\Bbb{R}$ is compact. Since $\Bbb{R}$ is a metric space, $S$ is also sequentially compact. In particular, a sequence $\{ x_n \}$ of points in $S$ such that $x_{n+1} \ge x_n$ has a subsequence that converges in $S$ and it must be the supremum. Is this a right approach to the sequence proof you mentioned? – Niki Di Giano Oct 28 '18 at 9:32